Moderate -0.8 This is a straightforward application of the parametric differentiation formula dy/dx = (dy/dt)/(dx/dt). Both derivatives are simple (1 ± 1/t), requiring only basic differentiation of logarithms, then substitution of t=2. It's more routine than average, involving direct recall of a standard technique with no problem-solving or conceptual challenges.
2 A curve is defined parametrically by the equations
$$x = t - \ln t , \quad y = t + \ln t \quad ( t > 0 )$$
Find the gradient of the curve at the point where \(t = 2\).
2 A curve is defined parametrically by the equations
$$x = t - \ln t , \quad y = t + \ln t \quad ( t > 0 )$$
Find the gradient of the curve at the point where $t = 2$.
\hfill \mbox{\textit{OCR MEI C4 2006 Q2 [5]}}